Talk:Initial value theorem

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Is F supposed to be the Laplace transform of ƒ? If so, it should say so. This article is written in a generally rather verbally challenged style. Michael Hardy (talk) 04:49, 6 February 2009 (UTC)Reply

... nobody answered, so I've done some further editing. Michael Hardy (talk) 22:13, 2 March 2009 (UTC)Reply

Does anyone think a proof or examples should be added? Krazyman (talk) 17:53, 27 February 2012 (UTC)Reply

I really don't see the necessity to complicate the proof that much. The initial value theorem is only makes sense for the one-sided Laplace transform, which means that ${\displaystyle f(0^{-})}$ does not make much sense (the derivative may not even exist at ${\displaystyle 0}$, e.g., when using the Heaviside function).

Hence, one can pose ${\displaystyle \lim _{s\to +\infty }sF(s)=\lim _{s\to +\infty }\left[\lim _{\epsilon \to 0^{+}}\int _{\epsilon }^{+\infty }{\frac {\partial }{\partial x}}f(x)e^{-sx}dx+f(\epsilon )\right]}$. By exchanging the limit and the summation (integral), which is allowed because of the uniform convergence for ${\displaystyle x>\epsilon >0}$, one obtains that the integral vanishes whenever ${\displaystyle \Re (s)>0}$ (implied by ${\displaystyle s\to +\infty }$), and thus the required answer.

Did I miss something essential here ?

ikingut 16:59, 26 November 2014 (UTC)